#include<bits/stdc++.h>

using namespace std;
using ll = long long;
#define endl '\n'

//子集空间O(2^n)
int main() {
    int a[] = {1, 2, 3, 4, 5, 6};
    int k = 10;
    //{1,2,3,4}  1111000
    int n = sizeof(a) / sizeof(a[0]);
    for (int i = 0; i < (1 << n); i++) {
        for (int k = n - 1; k >= 0; k--) {
            int bk = (i >> k) & 1;
            cout << bk;
        }
        cout << endl;
    }
    return 0;
}
